Question

joint probability density function f(x,y)=6(1-y) for 0<=x<=y<=1, and =0 elsewhere.
Find P(y<=1/2|x<=3/4).

Answer 1

I believe it would just be: \[
\int_{-\infty}^{1/2}\int_{-\infty}^{3/4}f(x,y)~dx~dy
\]

Answer 2

You can change the lower bounds to \(0\) since \(\int_{-\infty}^0\) will be \(0\).

Answer 3

Wait hold on. this is conditional probability?

Answer 4

Yes, it is

Answer 5

Well, what I do gives the probability that both are under their respective value.

Answer 6

So we need to use: \[
\Pr(Y|X) = \frac{\Pr(Y\cap X)}{\Pr(X)}
\]

Answer 7

We have the \(Y\cap X\) part down.

Answer 8

yes, but what next?

Answer 9

We need to find \(\Pr(X)\). I believe it would be given by \(f_X(x<3/4)\).

Answer 10

Do you know how to find that?

Answer 11

f_x(x)=3(1-x)^2

Answer 12

Do you sort of understand where I am getting these things?

Answer 13

I think you might have written \(f(x,y)\) wrong, because it has no \(x\).

Answer 14

not quite. I wrote it correctly. before this question, i had to find the marginal density functions for X and Y. they are f(x)=3(1-x)^2, f(y)=6y(1-y). now I have to find the conditional probability shown above. it is my "weak place" Textbook examples age giving examples with X or Y given, sort of P(X<=1|Y=2)

Answer 15

Hmm, now that I think about it, the \(x\leq y\) part does make things a bit tricky.

Answer 16

i think so

Answer 17

I guess you had to use \(f_X (x)= \int_x^1f(x,y)~dy\)
And then \(f_Y(y) = \int_0^y f(x,y)~dx\)

Answer 18

I used it to find f(x) and f(y). Now i need to use somehow f(x,y)/f(x)

Answer 19

If we assume that \(x\leq 3/4\)...

Answer 20

actually, that really doesn't help us much at all, lol.

Answer 21

But I'm not quite sure that is correct either.

Answer 22

Okay, now I'm thinking:\[
\int_0^{3/4}\int_x^{1/2}f(x,y)~dy~dx
\]

Answer 23

Hmmm, I think that ought to keep us so that \(x\leq y\)... don't you?

Answer 24

But maybe not, since we are letting \(x\) to up to \(3/4\)...

Answer 25

I don't know. I know the answer (32/63) but i have no idea how to get it.

Answer 26

Okay, I have another idea... what if we have \(y = t+x\)

Answer 27

and?

Answer 28

Then \(0\leq t\leq1-x\)

Answer 29

I am just trying to see if we can change this dependency on \(x\).

Answer 30

So we change: \[
\int_0^{3/4}\int_x^{1/2}f(x,y)~dy~dx =\int_0^{3/4}\int_0^{1/2-x} f(x,t+x)~dt~dx
\]

Answer 31

for me, it still looks like P(x,y) or P(x,t)